SOLUTION: This is my problem What is the "Vertex" & Sketch its graph y = -3x^2 + 6x Thanks, Paul

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Question 23313: This is my problem
What is the "Vertex" & Sketch its graph
y = -3x^2 + 6x
Thanks,
Paul
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
There are several ways to find the vertex of this. I think the easiest way is to use a formula for the vertex that comes from the quadratic formula. Remember that? x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ Well, this formula is just the first part of this without the radical. It turns out the vertex of a parabola y+=+ax%5E2+%2Bbx+%2Bc (which opens up or down!) is always at x=-b%2F%282a%29+

In your case, y+=+-3x%5E2+%2B+6x, a=-3 and b=6, so x=+-6%2F%282%28-3%29%29+=+1.

Now, if x=1, then
y+=+-3x%5E2+%2B+6x
y=-3+%2B6=+3

Vertex is at (1,3).

Check it out with the graph:
graph+%28500%2C500%2C+-10%2C10%2C-10%2C10%2C+-3x%5E2+%2B+6x%29+
Does this graph look like the vertex is at (1,3), and does the graph open down?

R^2 at SCC.